November 3, 2009

Multi-Step Inequalities CA SS: 5.0

p.184 (1-19:odd)

Per. 4 only
p.195 (2,4,6)

14 comments:

Nicholas L. Period 4 said...

How do you do problem #2 for Period 4?

DaisukeHper3 said...

Are we the only two that is in this?

Tariq M. Period 4 said...

i need help too.

Natasha L Per.4 said...

umm number 2.... so there's like 2 consultants i think.... so one got 500 more/less than the other one (it doesn't matter really, more/less,it's the same both ways...)

so this is how you set it up...

c+(c+500)=18,500
c=the one consultant that gets 500 less
so solve!!!

2c+500=18,500 - combine like terms

2c+500=18,500
-500 -500

2c=18,000
-- ---
2 2

c=9,000
c+500=9,500

so the two pays are $9,000 & $9,500
i think......
done!!!!

Tariq M. Period 4 said...

is anyone on? :(

Tariq M. Period 4 said...

no no #2 not the second problem the first problem

Joshua L. Per. 4 said...

i need help with #2 too... :(

Natasha L Per.4 said...

so you want the 1st problem??? wait wha???

Natasha L Per.4 said...

number one is kinda hard to explain

_______ ________ _______


you used one number that is not 0 or 1 in the first slot, there are a total of 4 possibilities for that single slot

so there are 5 possible numbers to the next slot, because you used one number on the first slot, so there are only a number of 5 possibilities left for the second slot.

Finally then you have 4 numbers for the last slot because you've used 2 numbers already for both the first and second slot

so....

4 x 5 x 4

meaning 4 possibilities times five possibilities times 4 possibilities, these number are based on haw many number of possibilities are available for each slot

so..... multiply!!!

4 x 5 x 4
20 x 4
80 possibilities!!! - i think

Natasha L Per.4 said...

haha i thought you meant the second problem on the assignment not the actual problem number, nikki and tariq sorry >.<

Eric X. Per. 3 TA said...

Number 2:

Since 0 or 1 can't be in the first digit of the prefix, there are only four other possibilties: 2,3,7, or 9.

Then the second digit will have five possibilites: 0, 1 , and four possibilties -1 because one of the numbers mentioned before would be in the first digit.

For example, if 2 was the first digit, then the second digit could be 0, 1, 3, 7, or 9, which is five possibilities.

Then the last digit would have four possibilities.

For example, If the first two digits were 2 and 3, then the last digit could be 0, 1, 7, or 9, which is four possibilities.

Multiply the 4 possibilities for the first digit, then the 5 for the second, and the 4 for the third. (4)(5)(4)=80

So I THINK there are 80 possibilities.

Natasha L Per.4 said...

me and eric got the same answer so it SHOULD not IS right......

Nicholas L. Period 4 said...

Thanks

Anonymous said...

whats the ? i can help!
but i need the actual equation
cuz im not in per 4